A ball is thrown from the location ( x 0 , y 0 )=(0,0) of a horizontal playground with an initial speed v 0 at an angle θ 0 from the +x -direction. The ball is to be hit by a stone, which is thrown at the same time from the location (x 1 , y 1 )=(L, 0) . The stone is thrown at an angle ( 180-θ 1 ) from the +x -direction with a suitable initial speed. For a fixed v 0 , when (θ 0 , θ 1 )= (45 ° , 45 ° ) , the stone hits the ball after time T 1 , and when (θ 0 , θ 1 )= (60 ° , 30 ° ) , it hits the ball after time T 2 . In such a case, (T 1 / T 2 ) 2 is \ \ \ \ \

Physics · Motion in Plane

JEE Advanced 2024 — Paper 2 — Question 20

A ball is thrown from the location $\left(\mathrm{x}_{0}, \mathrm{y}_{0}\right)=(0,0)$ of a horizontal playground with an initial speed $v_{0}$ at an angle $\theta_{0}$ from the $+x$ -direction. The ball is to be hit by a stone, which is thrown at the same time from the location $\left(x_{1}, y_{1}\right)=(L, 0)$ . The stone is thrown at an angle ( $180-\theta_{1}$ ) from the $+x$ -direction with a suitable initial speed. For a fixed $v_{0}$ , when $\left(\theta_{0}, \theta_{1}\right)=\left(45^{\circ}, 45^{\circ}\right)$ , the stone hits the ball after time $\mathrm{T}_{1}$ , and when $\left(\theta_{0}, \theta_{1}\right)=\left(60^{\circ}, 30^{\circ}\right)$ , it hits the ball after time $\mathrm{T}_{2}$ . In such a case, $\left(T_{1} / T_{2}\right)^{2}$ is $\_\_\_\_\_$

Answer: 2

Numerical answer — enter this value.

Step-by-step solution

$\quad \mathrm{v}_{0} \sin 45^{\circ}=\mathrm{v}_{1} \sin 45^{\circ}$

\Rightarrow \mathrm{v}_{1}=\mathrm{v}_{0} \quad \ldots \text { (I case) }

\mathrm{v}_{0} \sin 60^{\circ}=\mathrm{v}_{2} \sin 30^{\circ} \quad \Rightarrow \mathrm{v}_{2}=\mathrm{v}_{0} \sqrt{3} \ldots \text { (II case) }

T_{1}=\frac{L}{v_{1} \cos 45^{0}+v_{0} \cos 45^{\circ}}=\frac{L}{v_{0}(\sqrt{2})} \quad \Rightarrow T_{1}^{2}=\frac{L^{2}}{2 v_{0}^{2}}=v_{0} \sqrt{3}

$T_{2}=\frac{L}{v_{2} \cos 30^{\circ}+V_{0} \cos 60^{\circ}}$ $=\frac{L}{\left(v_{0} \sqrt{3}\right) \cdot \frac{\sqrt{3}}{2}+\frac{v_{0}}{2}}=\frac{2 L}{4 v_{0}}=\frac{L}{2 v_{0}}$ $\Rightarrow \quad \mathrm{T}_{2}^{2}=\frac{\mathrm{L}^{2}}{4 \mathrm{v}_{0}^{2}}$ $\left(\frac{T_{1}}{T_{2}}\right)^{2}=2$

Answer key and solution verified before publishing.

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Exam: JEE Advanced 2024
Paper: Paper 2
Subject: Physics
Chapter: Motion in Plane
Topic: Relative motion in Two Dimension